Question

It was reported that 30% of all credit card users in the U.S. pay their card card bill in full every month.

A random sample of size n=84 credit cards users is chosen. Which of the following is true regarding the effect of the increased sample size (from 21 to 84) on the mean and standard deviation of sampling distribution of pˆp^?

Answer 1

Solution

Let p = Proportion of all credit card users in the U.S. who pay their card bill in full every month.

And pcap = sample proportion of credit card users in the U.S. who pay their card bill in full every month.

Then, pcap ~ B(n, p), where n = sample size and p is as defined above……………………… (1)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success and x = number of successes, then

Mean (average) of X = E(X) = np………………………………………….............……..(2)

Variance of X = V(X) = np(1 – p)…………………………………………………………..(3)

Standatd Deviation of X = SD(X) = √{np(1 – p)} ………………………………………...(4)

Since proportion, p = X/n, Mean (p) = p and Variance (p) = p(1 - p)/n …………….. (5)

Now to work out the solution,

Given, 30% of all credit card users in the U.S. pay their card card bill in full every month, p = 0.3.

So, vide (1) and (5), for n = 21, Mean of sampling distribution of pcap = 0.3 and

Standard deviation of pcap = √{(0.3 x 0.7)/21} = 0.1.

When n = 84, Mean (pcap) = 0.3, but Standard deviation of pcap = √{(0.3 x 0.7)/84} = 0.05.

Thus, increasing sample size from 21 to 84 does not alter the mean, but halves the standard deviation. ANSWER

[Going beyond,

Mean does not depend on n and hence no change. Standard deviation has n in the denominator under square root. Since 84 is 4 times 21, the scaling down effect will be 2, i.e., halving.]

DONE